Can we define magnetic field at a point as:

Force on a point magnetic north pole at that point divided by its pole strength.

Anything wrong in this definition?

(The concept of point magnetic pole is an idealization as point magnetic poles don’t exist in nature and also there are no magnetic monopoles observed in nature.)

Question edited:

So does the above definition mean $\vec{B}$ or $\vec{H}$?

I think my definition does not represent $\vec{H}$ because we have not divided $\mu_0$ anywhere. So it represents $\vec{B}$ in free space. Am I correct?

  • $\begingroup$ My computer won't let me vote on questions, but I would upvote this one if I could. It's conceptual and, in my opinion, interesting. $\endgroup$ – Philip Wood Aug 30 '18 at 18:10

Your definition is exactly that of magnetic field strength, H, on the old cgs system. The oersted was its unit, a magnetic field strength of 1 dyne per unit pole. [A unit pole was such that if two unit poles were placed 1 cm apart in vacuo, there would be a force of 1 dyne between them.]

The study of electromagnetism was built on this foundation, and produced the edifice of electromagnetic theory that we have today – even though we may now choose to define things differently. So one answer to your question would be that there's nothing wrong with the 'per unit pole' definition. It has delivered the goods.

Another view is that it's odd to use the non-existent monopole as the basis for electromagnetism. In fact the subject used to be taught using long ball-ended (dumb-bell shaped) magnets. The ball at the 'North' (or South) end produced a magnetic field in the surrounding air radially outwards (or inwards) varying with an inverse square law. So the balls seemed to be behaving as monopoles. But there's a very important caveat… The net magnetic flux from the North pole ball (unlike from a North monopole) is zero! This is because as much flux approaches the ball through the 'bar' of the magnet as leaves through the air, and as much flux leaves the South pole ball through the bar of the magnet as enters through the air. One of the consequences is that we have to be especially careful studying magnetic materials using this approach. Arguably an approach based on Ampère's current loops or the corresponding quantum concept is less contrived.

| cite | improve this answer | |
  • $\begingroup$ I've added to my answer. $\endgroup$ – Philip Wood Aug 30 '18 at 16:01
  • $\begingroup$ So this measures H and not B? $\endgroup$ – lalala Aug 30 '18 at 17:02
  • 1
    $\begingroup$ Yes, that's right. In a vacuum physicists no longer consider that B and H are measuring different things, so it's quite nice that in the cgs system, a field of strength 1 oersted has a flus density of 1 gauss, (whereas in the SI, B⃗ =μ0H⃗, in which μ0 is a constant with units). The distinction between B and H is important, though, in materials with a relative permeability significantly different from 1. $\endgroup$ – Philip Wood Aug 30 '18 at 17:41
  • $\begingroup$ So does this prescription measure H in materials? $\endgroup$ – lalala Aug 31 '18 at 7:55
  • $\begingroup$ I believe so. I'd like to make sure but I no longer have any cgs (magnetic pole based) textbooks. $\endgroup$ – Philip Wood Aug 31 '18 at 8:53

Instead of magnetic North Pole, you could refer to a moving charge with a known velocity as a test particle. This avoids the issue of monopoles you mention.

| cite | improve this answer | |
  • $\begingroup$ I know that. Anyway in terms of magnetic poles, is the above stated definition valid? $\endgroup$ – N.G.Tyson Aug 29 '18 at 17:55
  • $\begingroup$ I don't know. It sounds like a simple analog of the electric field, but I don't know how to imagine the magnetic pole. $\endgroup$ – Jasper Aug 29 '18 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.